In Exercises 51–66, find
a. (fog) (x)
b. (go f) (x)
f(x)=4x-3, ...

Question

In Exercises 51–66, find 
a. (fog) (x)
b. (go f) (x) 
f(x)=4x-3, g(x) = 5x² - 2

Accepted Answer

So this problem says to solve F of G of X and G F F of X where f of X is equal to X plus square root of X and G of X is equal to X squared. Let's first take care of F of G F X. So F of G of X can be written with F on the outside and G on the inside of this function. So we're gonna take what we get for G of X and plug it in for our ffx. Look here half of X. We replace that with G F X. That means F of X squared. No, if I substitute this into our equation, we have an X there that sends the X squared and the plus squared of X plus squared of X squared. I'm gonna simplify this. So X squared is going to come down then we have a square root of X squared which is just X. So our F of G of X is equal to X squared plus X. Now we can look at our G F X. So we want G of F of X. Now that is G for athletics, which is X plus squared of X. Been going to substitute this to our equation we get equals two X plus squared of X squared. Now we can simplify this, we'll find out our X plus squared of X there plus squared of X. We can multiply these together by foiling. So we take our first terms multiply them outer terms, multiply in our terms and last terms and add all those modifications together. So we get X times X, which is X squared X sine squared of X is just X squared of X. On the inside we have another X squared of X and finally squared of X some squared of X is just X by itself. This simplifies them further with combining like terms two X squared, we have to export of excess that's too exclusive X. And finally just plus X. So our F of G of X is X squared plus X. And our G F F of X is X squared plus two X squared of X plus X, meaning the answer to our problem is answer D. Okay, I hope that you solve the problem. Thank you for watching. Goodbye.

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In Exercises 51–66, find a. (fog) (x) b. (go f) (x) f(x)=4x-3, g(x) = 5x² - 2